I found such problem while I was studying to my exam:limx→∞{∫x23x2{t⋅sin⁡{2t}dt}x2}

Since t⋅sin⁡2t→2− for t→∞, for each ϵ&gt;0 we can choose x0 such that, for t≥x0, we have 2−ϵ≤t⋅sin⁡2t≤2Thus for x≥x0(2−ε)2x2=∫x23x2(2−ε)dt≤∫x23x2t⋅sin⁡2tdt≤∫x23x22dt=4x2and(2−ε)2x2x2=2(2−ε)≤∫x23x2{t⋅sin⁡{2t}dt}x2≤4x2x2=4.The limit is 4.

Solved: "I found such problem while I was studying..."

High School
Calculus and Analysis
Calculus 1
Limits and continuity

Aryan Salinas

Answered question

2022-03-29

I found such problem while I was studying to my exam:

lim_{x \to \infty} {\frac{\int_{x^{2}}^{3 x^{2}} {t \cdot \sin {\frac{2}{t}} d t}}{x^{2}}}

Answer & Explanation

Abdullah Avery

Beginner2022-03-30Added 19 answers

Since

t \cdot \sin \frac{2}{t} \to 2^{-}

for

t \to \infty

, for each

ϵ > 0

we can choose

x_{0}

such that, for

t \geq x_{0}

, we have

2 - ϵ \leq t \cdot \sin \frac{2}{t} \leq 2

Thus for

x \geq x_{0}

(2 - ε) 2 x^{2} = \int_{x^{2}}^{3 x^{2}} (2 - ε) d t \leq \int_{x^{2}}^{3 x^{2}} t \cdot \sin \frac{2}{t} d t \leq \int_{x^{2}}^{3 x^{2}} 2 d t = 4 x^{2}

and

(2 - ε) \frac{2 x^{2}}{x^{2}} = 2 (2 - ε) \leq \frac{\int_{x^{2}}^{3 x^{2}} {t \cdot \sin {\frac{2}{t}} d t}}{x^{2}} \leq \frac{4 x^{2}}{x^{2}} = 4 .

The limit is 4.

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